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3 years ago

What's the relationship between energy and time

Engineering

1 answer:

boyakko [2]3 years ago

8 0

Answer:

The relationship between power, energy, and time can be described by the following equation : P = Δ E s y s Δ t. P is the average power output, measured in watts (W) ΔEsys is the net change in energy of the system in joules (J) - also known as work. Δt is the duration - how long the energy use takes - measured in seconds (s).

Explanation:

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The driveshaft of an automobile is being designed to transmit 238 hp at 3710 rpm. Determine the minimum diameter d required for

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Explanation:

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3 years ago

Air enters the compressor of a cold air-standard Brayton cycle with regeneration and reheat at 100 kPa, 300 K, with a mass flow

yanalaym [24]

Answer:

a. 47.48%

b. 35.58%

c. 2957.715 KW

Explanation:

$T_2 =T_1 + \dfrac{T_{2s} - T_1}{\eta _c}$

T₁ = 300 K

$\dfrac{T_{2s}}{T_1} = \left( \dfrac{P_{2}}{P_1} \right)^{\dfrac{k-1}{k} }$

$T_{2s} = 300 \times (10) ^{\dfrac{0.4}{1.4} }$

$T_{2s}$ = 579.21 K

T₂ = 300+ (579.21 - 300)/0.8 = 649.01 K

T₃ = T₂ + $\epsilon _{regen}$ (T₅ - T₂)

T₄ = 1400 K

Given that the pressure ratios across each turbine stage are equal, we have;

$\dfrac{T_{5s}}{T_4} = \left( \dfrac{P_{5}}{P_4} \right)^{\dfrac{k-1}{k} }$

$T_{5s}$ = 1400× $\left( 1/\sqrt{10} \right)^{\dfrac{0.4}{1.4} }$ = 1007.6 K

T₅ = T₄ + ( $T_{5s}$ - T₄)/ $\eta _t$ = 1400 + (1007.6- 1400)/0.8 = 909.5 K

T₃ = T₂ + $\epsilon _{regen}$ (T₅ - T₂)

T₃ = 649.01 + 0.8*(909.5 - 649.01 ) = 857.402 K

T₆ = 1400 K

$\dfrac{T_{7s}}{T_6} = \left( \dfrac{P_{7}}{P_6} \right)^{\dfrac{k-1}{k} }$

$T_{7s}$ = 1400× $\left( 1/\sqrt{10} \right)^{\dfrac{0.4}{1.4} }$ = 1007.6 K

T₇ = T₆ + ( $T_{7s}$ - T₆)/ $\eta _t$ = 1400 + (1007.6 - 1400)/0.8 = 909.5 K

a. $W_{net \ out}$ = cp(T₆ -T₇) = 1.005 * (1400 - 909.5) = 492.9525 KJ/kg

Heat supplied is given by the relation

cp(T₄ - T₃) + cp(T₆ - T₅) = 1.005*((1400 - 857.402) + (1400 - 909.5)) = 1038.26349 kJ/kg

Thermal efficiency of the cycle = (Net work output)/(Heat supplied)

Thermal efficiency of the cycle = (492.9525 )/(1038.26349 ) =0.4748 = 47.48%

b. $bwr = \dfrac{W_{c,in}}{W_{t,out}}$

bwr = (T₂ -T₁)/[(T₄ - T₅) +(T₆ -T₇)] = (649.01 - 300)/((1400 - 909.5) + (1400 - 909.5)) = 35.58%

c. Power = 6 kg *492.9525 KJ/kg = 2957.715 KW

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Answer:

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3-4=Irreversible expansion or throttling in which enthalpy remains constant

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What's the relationship between energy and time​

What's the relationship between energy and time